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#### renyikouniao

##### Member

- Jun 1, 2013

- 41

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- Jun 1, 2013

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- Jan 26, 2012

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$$ \frac{1}{1-(-1)} \int_{-1}^{1}(-x^{2}+1) \, dx= \int_{0}^{1}(-x^{2}+1) \, dx

= \frac{2}{3},$$

but

$$f\left( \frac{1+(-1)}{2}\right)= f(0) = 1 > \frac{2}{3} .$$

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- Jun 1, 2013

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awesome explaination!thank you.

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- Feb 7, 2012

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Perhaps the question should have read "Show that if f is a continuous concave downward function on [a,b],then the average of the function f is greater than [f(a)+f(b)]/2]." That is true, and you see why geometrically, if you notice that the graph of f lies above the line joining the points (a,f(a)) and (b,f(b)). Thus the area under the graph is greater than the area of the trapezium with vertices (a,0), (a,f(a)), (b,f(b)), (b,0). In other words, $$\int_a^bf(x)\,dx > \tfrac12(f(a)+f(b)(b-a),$$ from which $$\frac1{b-a}\int_a^bf(x)\,dx > \tfrac12(f(a)+f(b)).$$